HW06-T-2023S-Solutions

MEEN 260 Introduction to Engineering Experimentation Homework 6: Sensors and Strain Gages Assigned: February 28, 2023 Due: March 5, 2023, 11:59 PM General Instructions: Follow the posted Homework Guidelines when completing your assignment. You may scan your hand-written pages, or use a tablet writing device to convert your document to PDF format. Note that each problem must have: • A brief restatement of the problem in your own words (i.e. NOT a screen image of the problem statement) • A clear, legible, organized solution in symbolic form. Include any references to data or values used in your calculations. • Final answers identified (boxed, underlined) with appropriate significant digits and units Solid Mechanics Background: We will use some basic stress-strain principles in this homework assignment. The following definitions and equations are needed: Normal Stress, s : Force per unit area due to internal (normal direction) force within a mechanical component. By convention, positive stress (+) indicates tension, and negative stress (-) compression. The relationship between normal force F , cross-section area A , and normal stress s , is: ࠵? = ࠵? ࠵? Normal Strain (longitudinal/axial), e : The fractional change in the overall length of the component in the normal direction due to normal stress. By convention, positive strain (+) indicates elongation, and negative stress (-) indicates contraction. Normal strain is a dimensionless fraction of the unstressed dimension. A common unit of normal strain is the ‘micro-strain’, 1 μ -strain = fractional change of 10 -6 . E.g. A bar that has an original length of 1 m that increases in length by 1 μ m (overall length now 1.000001 m) has experienced 1 μ -strain (positive). The normal stress-strain relationship is given by Hooke’s Law with Young’s Modulus E , ࠵? = ࠵?࠵? Lateral/Transverse Normal Strain, e lat : An object that experiences longitudinal strain will also experience a complimentary lateral strain perpendicular to the longitudinal direction. The lateral strain can be computed from longitudinal strain using Poisson’s Ratio, n , a property of the material: ࠵? !"# = −࠵? ∙ ࠵? The negative sign indicates lateral strain is always opposite in sign relative to the longitudinal strain.

Problem #1: An aluminum ( E = 70 x 10 9 Pa, ࠵? = 0.33) rod supports a mass of 60 kg so that a uniform positive normal stress is created along the length of the rod. The rod has unstressed dimensions of length L = 0.3 m and diameter D = 5 mm. Strain gages having a gage factor of 1.8 and nominal (unstrained) resistance of 100 ohms can be placed at any of four possible positions on the aluminum rod (G1-G4) as indicated on the diagram. G1 and G2 orientations respond to longitudinal/axial strain, and the G3 and G4 orientations respond to lateral/transverse strain. The normal force caused by the weight: ࠵? $ = ࠵? = ࠵?࠵? = 60 ࠵?࠵? ∙ 9.8077 ࠵? ࠵? % 6 = 588.462 ࠵? The normal stress is: ࠵? = ࠵? ࠵? = 4࠵? ࠵?࠵? % = 4 ∙ 588.426 ࠵? ࠵?(0.005 ࠵?) % = 29.97 ࠵?࠵?࠵? The longitudinal strain measured by G1 and G2 is: ࠵? = ࠵? ࠵? = 29.97 ࠵?࠵?࠵? 70000 ࠵?࠵?࠵? = 428.14 ∙ 10 &’ = 428.14 ࠵?࠵?࠵?࠵?࠵?࠵?࠵? The transverse strain is determined by Poisson’s ratio: ࠵? #(")*+,(*, = −࠵? ∙ ࠵? !-). = −0.33 ∙ 428.14 ࠵?࠵?࠵?࠵?࠵?࠵?࠵? = −141.29 ∙ 10 &’ = −141.29 ࠵?࠵?࠵?࠵?࠵?࠵?࠵? The change in resistance for the G1/2 and G3/4 strain gages can be determined: ∆࠵? ࠵? = ࠵? ∆࠵? ࠵? = ࠵?࠵? ∆࠵? = ࠵? ∙ ࠵?࠵? G1 G2 G4 G3

࠵? , = ࠵? , ∙ ࠵? #-#"! The current flow in the left leg (R1 and R2) can be found from Ohm’s law: ࠵? 0% = ࠵? , ∙ ࠵? 0% The current flow in the right leg (R3 and R4) can be found from Ohm’s law: ࠵? 23 = ࠵? , ∙ ࠵? 23 The voltage at the node between R1 and R2 can be found by the voltage drop from V e and the current in the left leg: ࠵? 0%&)-4, = ࠵? , − ࠵? 0% ∙ ࠵? 0% The voltage at the node between R3 and R4 can be found by the voltage drop from V e and the current in the right leg: ࠵? 23&)-4, = ࠵? , − ࠵? 23 ∙ ࠵? 23 The measurement voltage Vo is the difference between the middle node of right and left legs (+ side is connected to right leg node, so the measurement voltage is right-node – left node): ࠵? - = (࠵? 23&)-4, ) − (࠵? 0%&)-4, ) For each case below, compute the deflection voltage based on the gage/resistor combination. Note the sign orientation of the Output Voltage measurement. A) Quarter-bridge, option 1 Bridge Resistor Position Strain Gage Position/ Ballast R 1 G1 R 2 Ballast R 3 Ballast R 4 Ballast B) Quarter-bridge, option 2 Bridge Resistor Position Strain Gage Position/ Ballast R 1 Ballast R 2 Ballast R 3 G3 R 4 Ballast

C) Half-bridge, option 1 Bridge Resistor Position Strain Gage Position/ Ballast R 1 G1 R 2 Ballast R 3 G3 R 4 Ballast D) Half-bridge, option 2 Bridge Resistor Position Strain Gage Position/ Ballast R 1 G1 R 2 Ballast R 3 Ballast R 4 G4 E) Full-bridge, option 1 Bridge Resistor Position Strain Gage Position/ Ballast R 1 G1 R 2 G2 R 3 G3 R 4 G4 F) Full-bridge, option 2 Bridge Resistor Position Strain Gage Position/ Ballast R 1 G1 R 2 G3 R 3 G4 R 4 G2

The components taken together represent a force measurement system where the input quantity is force ( F ) and the output quantity is voltage ( V o ). Based on this information determine the following characteristics of the measurement system: a) the sensitivity (value and units), b) the resolution (value and units), c) the range (value and units), d) in a couple of sentences, describe how you would test for the time constant of the system, e) if you had more strain gages of the same type to use, in a couple of sentences explain how you would increase the sensitivity and give, if possible, the factor by which you could increase the sensitivity. a) The definition of sensitivity is ࠵? = ࠵?࠵?࠵?࠵?࠵?࠵?࠵? ࠵?࠵?࠵?࠵?࠵?࠵? For the force measurement system, the output is the deflection or measurement voltage of the bridge, V o . The input is the applied force, F . We can use the stress-strain relationship, and bridge relationship to find a function for V o = f(F). First, determine normal stress in the bar with side length, s . ࠵? = ࠵? ࠵? = ࠵? ࠵? % Next, determine longitudinal strain, e : ࠵?࠵? = ࠵? = 1 ࠵? ∙ ࠵? ࠵? % The deflection voltage for a simple quarter bridge is given, ࠵? - = S ࠵?࠵? 5 4 T ࠵? = S ࠵?࠵? 5 4 T ∙ 1 ࠵? ∙ ࠵? ࠵? % The sensitivity is now: ࠵? = ࠵?࠵? - ࠵?࠵? = S ࠵?࠵? 5 4 T ∙ ࠵? ∙ 1 ࠵? % = S 3 ∙ 9࠵? 4 T ∙ S 1 200 ࠵?࠵?࠵? T ∙ 1 (10 ࠵?࠵?) % = 3.3375 ∙ 10 &6 ࠵? ࠵? = 0.3375 ࠵?࠵?/࠵?࠵?

b) The resolution of the force-sensing instrument must be stated in terms of input quantity, force. Resolution of Force = Resolution of Voltage / Sensitivity = 2963 N, 2.96 kN c) There are two considerations for the range of the system: the yield strength of the bar, the range of the voltage measurement device. Consider first the yield strength of the bar: ࠵? 75,!4 = ࠵? 8"9 ࠵? % ࠵? 8"9 = ࠵? 75,!4 ∙ ࠵? % = (200 ࠵?࠵?࠵?) ∙ (10 ࠵?࠵?) % = 20 ࠵?࠵? The second limitation is imposed by the range of the voltage measurement, ± 100 mV. The upper limit force values that would cause the maximum voltage output can be found from sensitivity: ࠵? 8"9 = ࠵? 8"9 ∙ ࠵? = (100 ࠵?࠵?) [0.3375 ࠵?࠵? ࠵?࠵? \ = 296.3 ࠵?࠵? The usable range of the force sensor is the smaller of the two maximum force values. The range is ± 20 kN d) To determine the time constant of the system, we need a dynamic (time variant) input to the system. Two typical dynamic input types for system response evaluation are 1) an impulse (hit the system with hammer), or 2) step-input, suddenly applied force. In either case, it would be necessary to measure system output as a function of time. If first order response is expected, we we should see an exponential decay of voltage vs time as system output, and can fit the measured data to an exponential decay function using our regression techniques for modified equation. e) The sensitivity of the force sensor system could be increased by using more strain gages, and increasing the bridge factor. The for a full bridge we need two longitudinal gages and two transverse gages. This would make the bridge factor: ࠵?࠵?࠵?࠵?࠵?࠵? ࠵?࠵?࠵?࠵?࠵?࠵? = (2 + 2࠵?) = (2 + 2 ∙ 0.3) = 2.6 The sensitivity is directly related to the bridge factor, so the new sensitivity would be 2.6 times larger than the quarter-bridge sensitivity. This would benefit resolution of the system but would not increase the range since range is limited by the yield strength of the bar. Problem #3: The ability to measure a stem cell’s mass is important since it reveals information about how close it is to mitosis (cell division). A device that detects the cell as early in its cell cycle (when it is small) as possible would be of great impact. As a mechanical engineer, you wonder whether you can design a cell-mass sensor with a micromachined cantilever beam (equipped with a strain gage) to detect the mass of a cell sitting at its tip:

d) suggest modifications to the detection system in order to improve your resolution, i.e., minimum detectable cell mass. Several options exist to improve (reduce) minimum mass detection: 1) Use multiple strain gages. The bridge factor can be as large as ‘4’ for a cantilever beam (4x the sensitivity of a single gage bridge) 2) Decrease the resolution of the electrical response ( Q ) 3) Increase the gage factor of the strain gages ( K ) 4) Increase the supply voltage, ( V s ) 5) Use a more flexible beam, smaller modulus ( E ), smaller width ( w ), thinner ( t ) 6) As indicated in part a, place gages at base x = 0 *The strain gage output V o can be expressed by: ࠵? - = S ࠵?࠵? * 4 T ࠵? where V s is the input voltage and K is the gage factor.